Past exam of the mathematical methods I course of the University of Oxford / 2014 / 12

A real symmetric matrix has $n$ freely chosen diagonal entries and $n(n-1)/2$ freely chosen entries above the diagonal. The entries below are then fixed by symmetry. Consequently

An explicit basis isIndeed every real symmetric matrix $\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$ has the unique linear combination $a{E}_1+c{E}_2+b{E}_3$.

The cyclic property of the trace gives $\mathrm{tr}(AB)=\mathrm{tr}(BA)$, which is C1. The linearity of trace and matrix multiplication gives C2 and C3. Finally symmetry of $A$ giveswith equality exactly when every entry is zero. Thus C4 also holds. This is the Frobenius inner product restricted to the real vector space of symmetric matrices.

Apply the cyclic property of the trace and $B{B}^{-1}=I$:This proves similarity invariance of trace without requiring $A$ or $B$ to be symmetric.

Taking the trace of the proposed commutator identity would givea contradiction for positive integer $n$ over the real numbers. No such finite real matrices exist. This is the trace of a commutator obstruction; it also applies to complex finite matrices.

Multiplicativity of the determinant gives $⟨A,B⟩=\det A\det B$. Thus C1 holds. For $n\ge 2$, C2 fails: choosing $A=B=I$ and $\lambda =2$ gives $\det (2I)=2^n\ne 2$. The same choice $A=B=C=I$ disproves C3, since $\det (B+C)=2^n\ne 2$. The nonnegativity portion of C4 holds, because $\det (A^2)=(\det A{)}^2\ge 0$, but its positive-definiteness clause fails: a nonzero singular matrix that is symmetric, such as $\mathrm{diag}(1,0,\dots ,0)$, has zero self-pairing.

Therefore this is not an inner product for $n\ge 2$. For the exceptional case $n=1$, the determinant is the scalar entry and the pairing is ordinary multiplication of real numbers, so all four conditions hold. The determinant pairing is not an inner product in higher dimension because it measures volume scaling rather than a bilinear length geometry.